I Landau's inertial frame logic

AI Thread Summary
The discussion focuses on the derivation of the Euler-Lagrange equations in the context of free particles, emphasizing that the Lagrangian, denoted as L, does not depend on time. It is established that L must be a function of velocity v, which is independent of time, leading to the conclusion that the Euler-Lagrange equations can still be applied. Landau's approach introduces the transformation of L in the K' inertial frame, where he modifies L by substituting v with dr/dt. This raises a question about the validity of this substitution given the initial assumption that v and q are not time-dependent. Ultimately, the discussion clarifies that Landau's addition of a total time derivative does not alter the equations of motion, maintaining the integrity of the analysis.
gionole
Messages
281
Reaction score
24
I had an interesting thought.

Let's only look at the free particle scenario.

We derive euler lagrange even without the need to know what exactly ##L## is (whether its a function of kinetic energy or not) - deriving EL still can be done. Though, because in the end, we end up with such EL(##\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot q} = 0##), we see that ##L## couldn't have been a function of ##\dot q## which depends on ##t##, because if ##\dot q## depends on ##t##, euler lagrange couldn't be applied to it as EL derivates ##L## wrt to ##\dot q##.

So at this time, we know ##L## is a function of ##v## in which ##v## doesn't depend on ##t##.

Then Landau tries to come up with what ##L## is. in the ##K'## inertial frame, he shows that ##L' = L(v^2) + \frac{dL}{dv^2}2v\epsilon##. Everything is clear till now, but then he changes ##v## into ##\frac{dr}{dt}##. How can he do that if the initial assumption is that ##v## and ##q## are not a function of ##t## in ##L## ? (I know that adding total time derivative doesn't change EOM, but this question is not about this)
 
Last edited:
Physics news on Phys.org
I think I figured out the logic in my head.

By that, he doesn't say that ##L'## is a function of ##v, q## which depend on $t$ - he doesn't say this. He just shows that adding total time derivative doesn't change EOM.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top